# AS math example - Thanks

Find the numbers M and N such that:

5+4x -x^2 = m - (x - n)^2
for all real values of x.

If someone could work this out, Then ill show my workings and ask questions (its to do with the arrangements of N^2 after expanding brackets mainly.

ASMATHSHELPME

## Answers and Replies

AKG
Homework Helper
Do you know how to complete the square?

ax² + bx + c
= a[x² + (b/a)x] + c
= a[x² + (b/a)x + (b/2a)² - (b/2a)²] + c
= a[x² + (b/a)x + (b/2a)²] - a(b/2a)² + c
= a[x + (b/2a)]² + (c - b²/4a)

Complete the square for the left side of your equation, and it will look like the right side, but instead of m and n, you'll have numbers, and those are obviously the numbers you're trying to find.

Complete the square, to put the equation in the form, (x - n)^2 + m , and then you have your values for n and m.

multiply both sides by -1 , to get, x^2-4x-5 which can be written as (x-2)^2-9 , which is in the form (x - n)^2 + m

so n=2, m=-9

VietDao29
Homework Helper
roger said:
so n=2, m=-9
Recheck that, roger. The problem states: 5+4x -x^2 = m - (x - n)^2, not 5+4x -x^2 = m + (x - n)^2
Viet Dao,

whoops,

m=9